Find the minimum value of f(x)=2x3+3 in the interval [-2,2]. The absolute minimum value is...

Step 1
We first find the critical point by solving f'(x)=0
${\displaystyle f\left(x\right)=2x^3+3}$
${\displaystyle f^{\prime }\left(x\right)=6x^2}$
${\displaystyle 0=6x^2}$
${\displaystyle x^2=0}$
x=0
Step 2
Then we check the values of f(x) at the critical point and the endpoints
${\displaystyle f\left(x\right)=2x^3+3}$
At x=-2, f(-2)=-13
At x=0, f(0)=3
At x=2, f(2)=19
So the minimum value is -13.
Answer: The absolute minimum value is -13 at x=-2